Divisible sum pairs

#beginners #programming #tutorial #javascript

Prepare your favorite cup of coffee, because we are about to enter the fantastic world of Divisible sum pairs.

The problem

The solution

To start our solution, let's define the divisibleSumPairs function that will receive the following parameters:

n: length of array arr;
k: integer divisor;
ar: input array.

function divisibleSumPairs(n, k, ar) {}

Next we will initialize the counter variable with the value 0:

let counter = 0;

Now let's go through the ar array from the first element to the penultimate element:

for (let i = 0; i < n - 1; i++) {}

Inside the first loop, we will start a second loop, to cycle through the elements of the ar array from the i + 1 element to the last element:

for (let j = i + 1; j < n; j++) {}

After traversing the array ar with the two loops (integrating through the element i and the element j (i + 1)), we will check if the sum of the elements in the indices i and j of the array arr is divisible by k without remainder:

if ((ar[i] + ar[j]) % k === 0) {}

If this condition is met, we will increase the value of the counter variable:

counter++;

Thus, we will have the number of pairs that satisfy this condition counted in counter.

Finally, let's return the value of counter:

return counter;

Final resolution

After following the step by step we have our final resolution:

function divisibleSumPairs(n, k, ar) {
    let counter = 0;

    for (let i = 0; i < n - 1; i++) {
      for (let j = i + 1; j < n; j++) {
        if ((ar[i] + ar[j]) % k === 0) {
          counter++;
        }
      }
    }

    return counter;
}

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Top comments (8)

Tracy Gilmore • May 13 • Edited

Hi Klecianny,
I would like to offer two alternative implementations. Not better (probably worse), just different approaches. There is still a double pass through the array, just not as explicate as in your solution.

Using reduce & filter methods

  function divisibleSumPairs(k, ar) {
    return ar.reduce(
      (acc, ar_i, i) =>
        acc + ar.slice(i + 1).filter(ar_j => !((ar_i + ar_j) % k)).length,
      0
    );
  }

Using recursion with a filter method

  function divisibleSumPairs(k, ar) {
    return _divisibleSumPairs(0, ...ar);

    function _divisibleSumPairs(count, ar_i, ...ar_) {
      return ar_.length
        ? _divisibleSumPairs(
            count + ar_.filter(ar_j => !((ar_i + ar_j) % k)).length,
            ...ar_
          )
        : count;
    }
  }

Klecianny Melo • Jun 6

Hello @tracygjg, how are you? Thank you very much for this solutions for the current problem! Programming is fascinating because we can solve the same problem in different ways. I look forward to your interactions in future articles! 😁